Electromagnetic Waves, Polarization, and Photons – Step-by-Step Study Guidance

Q1. A radio antenna broadcasts a 1.0 MHz radio wave with 25 kW of power. Assume that the radiation is emitted uniformly in all directions.

• (a) What is the wave’s intensity 30 km from the antenna?

• (b) What is the electric field amplitude at this distance?

Background

Topic: Electromagnetic Waves – Intensity and Electric Field Amplitude

This question tests your understanding of how electromagnetic wave energy spreads out from a point source and how to relate intensity to the electric field amplitude.

Key Terms and Formulas

• Intensity ($I$): Power per unit area, $I = \frac{P}{A}$

• For a point source radiating uniformly in all directions, area $A = 4\pi r^2$ (surface area of a sphere)

• Relationship between intensity and electric field amplitude: $I = \frac{1}{2} c \varepsilon_0 E_0^2$

• Where:

  • $P$ = total power emitted (in watts)

  • $r$ = distance from the source (in meters)

  • $c$ = speed of light ($3.0 \times 10^8$ m/s)

  • $\varepsilon_0$ = permittivity of free space ($8.85 \times 10^{-12}$ C$^2$/N·m$^2$)

  • $E_0$ = electric field amplitude (in V/m)

Step-by-Step Guidance

1. Convert all quantities to SI units. For example, power $P = 25$ kW $= 25 \times 10^3$ W, and distance $r = 30$ km $= 3.0 \times 10^4$ m.

2. Calculate the intensity at 30 km using the formula for a point source:

   $I = \frac{P}{4\pi r^2}$

   Plug in the values for $P$ and $r$ (in meters).

3. To find the electric field amplitude $E_0$ at this distance, use the relationship:

   $I = \frac{1}{2} c \varepsilon_0 E_0^2$

   Rearrange to solve for $E_0$:

   $E_0 = \sqrt{\frac{2I}{c \varepsilon_0}}$

4. Substitute your calculated value of $I$ and the constants $c$ and $\varepsilon_0$ into the equation for $E_0$.

Try solving on your own before revealing the answer!

Final Answer:

(a) $I \approx 2.2 \times 10^{-6}$ W/m$^2$

(b) $E_0 \approx 0.041$ V/m

We used the formulas for intensity from a point source and the relationship between intensity and electric field amplitude to find these values.

Q2. The intensity of a polarized electromagnetic wave is 10 W/m2. What will be the intensity after passing through a polarizing filter whose axis makes the following angles with the plane of polarization?

• (a) $\theta = 0^\circ$

• (b) $\theta = 30^\circ$

• (c) $\theta = 45^\circ$

• (d) $\theta = 60^\circ$

• (e) $\theta = 90^\circ$

Background

Topic: Polarization of Light – Malus’s Law

This question tests your understanding of how the intensity of polarized light changes when passing through a polarizing filter at various angles.

Key Terms and Formulas

• Malus’s Law: $I = I_0 \cos^2 \theta$

• Where:

  • $I_0$ = initial intensity (before the filter)

  • $I$ = transmitted intensity (after the filter)

  • $\theta$ = angle between the light’s polarization direction and the filter axis

Step-by-Step Guidance

1. For each angle, use Malus’s Law to set up the calculation:

   $I = I_0 \cos^2 \theta$

2. Plug in $I_0 = 10$ W/m$^2$ and the given value of $\theta$ for each part (convert degrees to radians if using a calculator in radian mode).

3. Calculate $\cos \theta$ for each angle, then square the result to get $\cos^2 \theta$.

4. Multiply $I_0$ by $\cos^2 \theta$ to find the transmitted intensity for each case.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $I = 10$ W/m$^2$

• (b) $I = 7.5$ W/m$^2$

• (c) $I = 5.0$ W/m$^2$

• (d) $I = 2.5$ W/m$^2$

• (e) $I = 0$ W/m$^2$

These results follow directly from Malus’s Law for each angle.

Q3. A 50 mW laser beam is polarized horizontally. It then passes through two polarizers. The axis of the first polarizer is oriented at 30° from the horizontal, and that of the second is oriented at 60° from the horizontal. What is the power of the transmitted beam?

Background

Topic: Polarization – Multiple Polarizers and Malus’s Law

This question tests your ability to apply Malus’s Law sequentially for light passing through more than one polarizer.

Key Terms and Formulas

• Malus’s Law for each polarizer: $I = I_0 \cos^2 \theta$

• Power is proportional to intensity if the area is unchanged: $P \propto I$

• For two polarizers, apply Malus’s Law twice, using the angle between the polarization direction and each polarizer’s axis.

Step-by-Step Guidance

1. Start with the initial power $P_0 = 50$ mW. The beam is initially polarized horizontally.

2. After the first polarizer (axis at $30^\circ$ to the horizontal), use Malus’s Law:

   $P_1 = P_0 \cos^2(30^\circ)$

3. The light after the first polarizer is now polarized at $30^\circ$ from the horizontal. The second polarizer is at $60^\circ$ from the horizontal, so the angle between the new polarization direction and the second polarizer is $60^\circ - 30^\circ = 30^\circ$.

4. Apply Malus’s Law again for the second polarizer:

   $P_2 = P_1 \cos^2(30^\circ)$

   Or, combine both steps: $P_2 = P_0 \cos^2(30^\circ) \cos^2(30^\circ)$

5. Calculate $\cos(30^\circ)$, square it, and multiply by the initial power twice to find the transmitted power.

Try solving on your own before revealing the answer!

Final Answer:

$P_2 \approx 28$ mW

We applied Malus’s Law twice, once for each polarizer, using the correct angles between the polarization directions and the polarizer axes.

Q4. Three laser beams have wavelengths $\lambda_1 = 400$ nm, $\lambda_2 = 600$ nm, and $\lambda_3 = 800$ nm. The power of each laser beam is 1 W.

• (a) Rank in order, from largest to smallest, the photon energies $E_1$, $E_2$, and $E_3$ in these three laser beams. Explain.

• (b) Rank in order, from largest to smallest, the number of photons per second $N_1$, $N_2$, and $N_3$ delivered by the three laser beams. Explain.

Background

Topic: Photons – Energy and Number of Photons

This question tests your understanding of the relationship between photon energy, wavelength, and the number of photons emitted per second for a given power.

Key Terms and Formulas

• Photon energy: $E = \frac{hc}{\lambda}$

• Number of photons per second: $N = \frac{P}{E}$

• Where:

  • $h$ = Planck’s constant ($6.63 \times 10^{-34}$ J·s)

  • $c$ = speed of light ($3.0 \times 10^8$ m/s)

  • $\lambda$ = wavelength (in meters)

  • $P$ = power (in watts)

Step-by-Step Guidance

1. For part (a): Use $E = \frac{hc}{\lambda}$ to compare the photon energies for each wavelength. Remember, energy is inversely proportional to wavelength.

2. Rank the energies: The shorter the wavelength, the higher the photon energy.

3. For part (b): Use $N = \frac{P}{E}$ to compare the number of photons per second for each beam. Since power is the same, the beam with lower photon energy will emit more photons per second.

4. Rank the photon numbers: The longer the wavelength, the more photons per second are needed to deliver the same power.

Try solving on your own before revealing the answer!

Final Answers:

• (a) $E_1 > E_2 > E_3$ (shortest wavelength has the highest energy)

• (b) $N_3 > N_2 > N_1$ (longest wavelength emits the most photons per second for the same power)

This follows from the inverse relationship between photon energy and wavelength, and the direct relationship between photon number and wavelength for constant power.